java string stringBuffer problem

  
  
  public class Test{
  	void Trans(int intIndex, String str, StringBuffer strBuffer)
  	{
  		intIndex += 1;
  		str += '1';
  		strBuffer.append('1');
  	}
  	
  	
  	public static void main(String[] args) {
  		int intIndex = 100;
  		String str = "100";
  		StringBuffer strBuffer = new StringBuffer("100");
  		Test t = new Test();
  		t.Trans(intIndex, str, strBuffer);
  		System.out.println(intIndex);
  		System.out.println(str);
  		System.out.println(strBuffer);
  	}
  }
  

ask why the above output
100
100
1001

------ Solution ------------------------------------ --------
first parameter is the original data , is constant, does not change the value of the original address , the second is the final type , is constant , so the cost will not change , only the the third is a pass into the object object , so you can change its value. . .
------ Solution ---------------------------------------- ----
http://www. ibm.com / developerworks / cn / java / j-lo-processthread / index.html processes and threads do not understand the process


------ For reference only ---------------------------------- -----
intIndex is shaped relationship between the parameter and argument will not change
str is a string, immutable
strBuffer the append method behind the increase in the string string
------ For reference only ---------------------- -----------------
easiest to understand is that the first and the second argument is the value, and the third argument is the reference
------ For reference only ---------------------------------------
Another problem is ___________________
process is created and released , the scheduling process is actually _______________________ .
In java how to understand the process? ? Because much of the score , posted in this problem, right ?
------ For reference only -------------------------------------- -
because your Trans () method does not return a value, so
System.out.println (intIndex);
System.out.println (str);
output is 100 ,
because stringBuffer append a 1 , it will be initialized 100 stitching on the back of a string is 1001